# Thread: Absolute Convergence of Integral

1. ## Absolute Convergence of Integral

I'm having a hard time showing that $\lim_{c \to \0} \int_{c}^{1}|\frac{sin(\frac{1}{x})}{x}|dx$ fails to converge absolutely, any insight would be greatly appreciated.

2. Try writing $\displaystyle \sin{\left(\frac{1}{x}\right)}$ as a Taylor Series, then simplifying $\displaystyle \left|\frac{\sin{\left(\frac{1}{x}\right)}}{x} \right|$.

3. Another way: If you know that $\int_1^{\infty} \left| \frac{\sin (z)}{z} \right|dz$ is infinite, just apply a change of variables.

4. ## Thanks

Thanks to both of you for your help. I did try one way before I read this, maybe you can tell me this works...

$\lim_{c\to\0}\int_{c}^{1}|\frac{sin(\frac{1}{x})}{ x}|dx\geqslant \sum_{k = 1}^\infty\int_{\frac{2}{(4k+1)\pi} }^{\frac{2}{(4k-1)\pi} }|\frac{sin(\frac{1}{x})}{x}|dx \geqslant \sum_{k = 1}^\infty\int_{\frac{2}{(4k+1)\pi} }^{\frac{2}{(4k-1)\pi} }\frac{dx}{x}$

...and then just evaluate, I could be way off.

5. Originally Posted by eastman
Thanks to both of you for your help. I did try one way before I read this, maybe you can tell me this works...

$\lim_{c\to\0}\int_{c}^{1}|\frac{sin(\frac{1}{x})}{ x}|dx\geqslant \sum_{k = 1}^\infty\int_{\frac{2}{(4k+1)\pi} }^{\frac{2}{(4k-1)\pi} }|\frac{sin(\frac{1}{x})}{x}|dx \geqslant \sum_{k = 1}^\infty\int_{\frac{2}{(4k+1)\pi} }^{\frac{2}{(4k-1)\pi} }\frac{dx}{x}$

...and then just evaluate, I could be way off.
How did you obtain the last inequality? There is alway a zero in those intervals so the elementary minimum bound doesn't work here.

6. Sorry, I'm slightly confused, could you elaborate?

7. Disregard the second part of my last message, could you explain how you got the last inequality?

8. You're right the last inequality does not hold, sin vanishes over those intervals.. I ended up doing it with the change of variables like you orginally said. Thanks for your help.